Simplify and expand the following expression: $ \dfrac{1}{4r + 16}- \dfrac{5}{2r - 10}+ \dfrac{4}{r^2 - r - 20} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{1}{4r + 16} = \dfrac{1}{4(r + 4)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{5}{2r - 10} = \dfrac{5}{2(r - 5)}$ We can factor the quadratic in the third term: $ \dfrac{4}{r^2 - r - 20} = \dfrac{4}{(r + 4)(r - 5)}$ Now we have: $ \dfrac{1}{4(r + 4)}- \dfrac{5}{2(r - 5)}+ \dfrac{4}{(r + 4)(r - 5)} $ The least common multiple of the denominators is: $ 8(r + 4)(r - 5)$ In order to get the first term over $8(r + 4)(r - 5)$ , multiply by $\dfrac{2(r - 5)}{2(r - 5)}$ $ \dfrac{1}{4(r + 4)} \times \dfrac{2(r - 5)}{2(r - 5)} = \dfrac{2(r - 5)}{8(r + 4)(r - 5)} $ In order to get the second term over $8(r + 4)(r - 5)$ , multiply by $\dfrac{4(r + 4)}{4(r + 4)}$ $ \dfrac{5}{2(r - 5)} \times \dfrac{4(r + 4)}{4(r + 4)} = \dfrac{20(r + 4)}{8(r + 4)(r - 5)} $ In order to get the third term over $8(r + 4)(r - 5)$ , multiply by $\dfrac{8}{8}$ $ \dfrac{4}{(r + 4)(r - 5)} \times \dfrac{8}{8} = \dfrac{32}{8(r + 4)(r - 5)} $ Now we have: $ \dfrac{2(r - 5)}{8(r + 4)(r - 5)} - \dfrac{20(r + 4)}{8(r + 4)(r - 5)} + \dfrac{32}{8(r + 4)(r - 5)} $ $ = \dfrac{ 2(r - 5) - 20(r + 4) + 32} {8(r + 4)(r - 5)} $ Expand: $ = \dfrac{2r - 10 - 20r - 80 + 32}{8r^2 - 8r - 160} $ $ = \dfrac{-18r - 58}{8r^2 - 8r - 160}$ Simplify: $ = \dfrac{-9r - 29}{4r^2 - 4r - 80}$